Question

A coil having $N$ turns is wound tightly in the form of a spiral with inner and outer, radii $a$ and $b$ respectively. When a current $I$ passes through the coil, the magnetic field at its centre is:

• A. $\frac{{\mu }_{0}NI}{b}$
  
• B. $\frac{2{\mu }_{0}NI}{b}$
  
• C. $\frac{{\mu }_{0}NI}{2(b-a)}ln\left(\frac{b}{a}\right)$
  
• D. $\frac{{\mu }_{0}I}{2(b-a)}ln\left(\frac{b}{a}\right)$
  

Answer 1

The correct option is C $\frac{{\mu }_{0}NI}{2(b-a)}ln\left(\frac{b}{a}\right)$


Outer radius of spiral is $b$ and its inner radius $a$, thus the width of the spiral is $(b-a)$.

Number of turns of spiral per unit width $=\frac{N}{(b-a)}$

Consider an elemental ring of radius $x$ and thickness $dx$ as shown in the figure.

Number of turns in the elemental ring is,
$dN=\frac{Ndx}{(b-a)}$


Magnetic field at the centre due to the ring element is,

$dB=\frac{{\mu }_0\left(dN\right)I}{2x}$

$\Rightarrow dB=\frac{{\mu }_{0}I}{2x}\frac{Ndx}{(b-a)}$

Thus, the magnetic field at centre of spiral is,
$B=\int dB$

$\Rightarrow B=\int \frac{{\mu }_{0}IN}{2(b-a)}\frac{dx}{x}$

$\Rightarrow B={\int }_{x=a}^{x=b}\frac{{\mu }_{0}NI}{2(b-a)}\frac{dx}{x}$

$\Rightarrow B=\frac{{\mu }_{0}NI}{2(b-a)}[lnx{]}_a^b$

$B=\frac{{\mu }_{0}NI}{2(b-a)}[lnb-lna]$

$B=\frac{{\mu }_{0}NI}{2(b-a)}ln\left(\frac{b}{a}\right)$

Why this Question ? Key point - The magnetic field due to the entire spiral at the centre is not constant, so we need to choose an elementary ring of infinitesimally small thickness $dx$. The magnetic field due to elementary ring can be assumed to be constant at the centre. Finally, with the help of calculus, we can determine the magnetic field due to entire spiral.